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Let's

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say I have the indefinite integral 1 over the square

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root of 3 minus 2x squared.

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Of course I have a dx there.

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So right when I look at that, there's no obvious

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traditional method of taking this antiderivative.

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I don't have the derivative of this sitting someplace else in

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the integral, so I can't do traditional u-substitution.

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But what I can do is I could say, well, this almost looks

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like some trig identities that I'm familiar with, so maybe I

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can substitute with trig functions.

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So let's see if I can find a trig identity that

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looks similar to this.

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Well, our most basic trigonometric identity-- this

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comes from the unit circle definition-- is that the sine

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squared of theta plus the cosine squared of

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theta is equal to 1.

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And then if we subtract cosine squared of theta from both

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sides, we get-- or if we subtract sine squared of theta

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from both sides, we could do either-- we could get cosine

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squared of theta is equal to 1 minus sine squared of theta.

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We could do either way.

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But this, all of a sudden, this thing right here, starts to

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look a little bit like this.

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Maybe I can do a little bit of algebraic manipulation to make

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this look a lot like that.

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So the first thing, I would like to have a 1 here-- at

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least, that's how my brain works-- so let's factor out a

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3 out of this denominator.

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So this is the same thing as the integral of 1 over the

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square root of-- let me factor out a 3 out of this expression.

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3 times 1 minus 2/3x squared.

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I did nothing fancy here.

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I just factored the 3 out of this expression,

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that's all I did.

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But the neat thing now is, this expression looks a

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lot like that expression.

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In fact, if I substitute, if I say that this thing right here,

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this 2/3x squared, if I set it equal to sine squared theta, I

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will be able to use this identity.

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So let's do that.

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Let's set 2/3x squared, let's set that equal to sine

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squared of theta.

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So if we take the square root of both sides of this equation,

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I get the square root of 2 over the square root of 3 times x is

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equal to the sine of theta.

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If I want to solve for x, what do I get?

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And, well, we're going to have to solve for both

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x and for theta, so let's do it both ways.

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First, let's solve for theta.

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If we solve for theta, you get that theta is equal to the

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arcsine, or the inverse sine, of square root of 2 over

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square root of 3x.

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That's if you solve for theta.

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Now, if you solve for x, you just multiply both sides of

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this equation times the inverse of this and you get x is equal

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to-- divide both sides of the equation by this or multiply it

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by the inverse-- is equal to the square root of 3 over the

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square root of 2 times the sine of theta.

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And we were going to substitute this with sine squared of

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theta, but we can't leave this dx out there.

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We have to take the integral with respect to d theta.

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So what's dx with respect to d theta?

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So the derivative of x with respect to theta is equal to

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square root of 3 over square root of 2.

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Derivative of this with respect to theta is just cosine of

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theta, and if we want to write this in terms of dx, we could

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just write that dx is equal to square root of 3 over the

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square root of 2 cosine of theta d theta.

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Now we're ready to substitute.

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So we can rewrite this expression up here-- I'll do it

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in this reddish color-- I was using that, let me do

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it in the blue color.

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We can rewrite this expression up here now.

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It's an indefinite integral of-- dx is on the

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numerator, right?

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Instead of writing this 1 times dx, I could have

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just written a dx up here.

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That could be a dx just like that.

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You're just multiplying it times dx.

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So what's dx?

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dx is this business.

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I'll do it in yellow.

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dx is this right here.

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So it's the square root of 3 over the square root of

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2 cosine theta d theta.

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That's what dx was.

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Now, the denominator in my equation, I have the square

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root of 3 times-- now it's 1 minus.

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Now I said 2/3x squared is equal to sine squared of theta.

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Now how can I simplify this?

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Well, what's 1 minus sine squared of theta?

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That's cosine squared of theta.

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So this thing right here is cosine squared of theta.

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So my indefinite integral becomes the square root of 3

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over the square root of 2 cosine theta d theta, all of

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that over the square root of 3 times the cosine

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squared of theta.

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That just became cosine squared of theta.

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So let's just take the square root of this bottom part.

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So this is going to be equal to-- I'll do an arbitrary

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change of colors-- square root of 3 over the square root of 2

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cosine of theta d theta, all of that over-- what's the

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square root of this?

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It's equal to the square root of 3 times the square root

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of cosine squared, so times cosine of theta.

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Now, this simplifies things a good bit.

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I have a cosine of theta divided by a cosine of theta,

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those cancel out, so we'll just get 1, and then I have a square

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root of 3 up here divided by a square root of 3, so those two

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guys are going to cancel out, so my integral simplifies

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nicely to 1 over square root of 2 d theta.

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Or even better, I could write this-- this is just a constant

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term, I could take it out of my integral-- it equals 1 over the

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square root of 2 times my integral of just d theta.

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And this is super easy.

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This is equal to 1 over the square root of 2

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times theta plus c.

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Plus some constant.

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I mean, you could say that the integral of this is theta plus

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c and then you'd multiply the constant times this, but it's

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still going to be some arbitrary constant.

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I think you know how to take the antiderivative of this.

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But are we done?

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Well, no.

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We want to know our indefinite integral in terms of x.

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So now we have to reverse substitute.

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So what is theta?

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We figured that out here. theta is equal to arcsine the square

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root of 2 over the square root of 3x.

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So our original indefinite integral, which was all of this

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silliness up here, now that I reverse substitute for theta or

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put x back in there, it's 1 over the square root

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of 2 times theta.

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theta is just this, is just arcsine of square root of 2

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over square root of 3 x, and then I have this constant

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out here, plus c.

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So this right here is the antiderivative of

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1 over the square root of 3 minus 2x squared.

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So hopefully you found that helpful.

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I'm going to do a couple of more videos where we go through

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a bunch of these examples, just so that you get

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familiar with them.

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